Carrier wave amplitude control in power source for stabilizing dc voltage by utilizing frequency dependence of resonance

ABSTRACT

In the power supply composed of a driver circuit generating the carrier, a resonance circuit driven by the carrier and a rectification circuit generating the dc. voltage by rectifying the amplitude-modulated carrier supplied by the resonance circuit and stabilizing the output voltage by the feeding back the voltage error between the output voltage and a reference voltage externally supplied to set up the output voltage, to the frequency and the amplitude of the carrier, the frequency response of the dc. power supply is improved both by providing the pole located at the origin with a transfer function where the dc. voltage, generated by rectifying and smoothing the output of the resonance circuit, is fed back to the frequency of the carrier and by not providing the pole at the origin with a transfer function where the dc. voltage is fed back to the amplitude of the carrier.

TECHNICAL FIELD

The present invention relates to stabilization of the output voltage ofthe power supply which generates direct-current output voltage using aresonance circuit.

BACKGROUND ART

A power supply using a piezoelectric transformer is such a power supplythat generates voltage using a resonance circuit. Since power density ofthe piezoelectric transformer is several times larger than that of theconventional magnetic transformer and can be operated in such a range ofhigh frequency where the conventional magnetic transformer shows largeloss and becomes impractical, the piezoelectric transformer of the samerating in power can be made several times smaller in size than that ofthe magnetic transformer. But the piezoelectric transformer dependslargely on load and frequency, showing characteristics different fromthe conventional magnetic transformer, which prevent practical usage ofthe piezoelectric transformer in a power supply.

Such a stabilized direct current (dc.) voltage supply regulates itsoutput voltage using frequency dependence of amplitude ratio. Theamplitude ratio, being the voltage ratio of the input to the output ofthe piezoelectric transformer, shows resonance characteristics againstthe frequency of the carrier. From a view point of efficiency, thepiezoelectric transformer is supplied with such the carrier that ishigher in frequency than the resonance of the piezoelectric transformer.So, for example, the frequency of the carrier is lowered and moved tothe resonance to increase the output voltage. Yet the fall of thefrequency causes drop of the output voltage which is generated byrectifying the output carrier of the transformer. Namely, to lower thefrequency in order to increase the output voltage causes temporary dropof the output voltage. The temporary drop of the output voltage grows inamplitude as the load becomes heavy.

In the case that the carrier is higher than the resonance in frequencyin the voltage supply, the transfer function of the output voltage has azero in the right half plane as the function of the carrier frequency.It is difficult to implement a large loop gain for the circuit whosetransfer function has zeros in the right half plane because the circuithas a narrow range of parameters for stable operation. Furthermore it isdifficult to resolve the zeros in the right half plane by feedbackbecause the circuit having poles in the right half plane is not stable.

-   [Patent Citation 1]    1:    2002-359967-   [Patent Citation 2]    2:    2005-137085-   [Patent Citation 3]    3:    2006-130906-   [Patent Citation 4]    4:    2007-018715-   [Patent Citation 5]    5:    2007-101471-   [Patent Citation 6]    6: PCT/JP2007/000477

The patent reference 1 makes it a subject to offer the simple circuit ofthe dc. power supply providing the stabilized high voltage withsufficient efficiency. Efficiency is improved by using not aconventional electromagnetic transformer but a piezoelectrictransformer. The high voltage is stabilized by using the frequencydependability of the resonance characteristics of a piezoelectrictransformer. The piezoelectric transformer simplifies the circuit andreduces parts in number, by which the subject is solved.

The patent reference 2 realizes improvement of the output voltage bothin the accuracy of stabilization and in the speed of response byimplementing feedback of a little delay in addition to the feedback of alarge delay where the output voltage is stabilized based on thefrequency dependability of the resonance characteristic. The presentinvention is that an idea of the feedback in the patent reference 2 isapplied to the amplitude of the carrier.

The patent reference 3 concerns the stabilized dc. voltage power supplywhere stabilization is based on the frequency dependence of theresonance and gives composition and its constants of the feedbackcircuit stabilizing the dc. output voltage that is supplied to the loadof a wide range, where the transfer function feeding back the outputvoltage to the frequency of the carrier that drives the resonancecircuit has a pole located at the origin.

The patent reference 4 concerns the stabilized dc. voltage power supplywhere stabilization is based on the frequency dependence of theresonance and gives composition and its constants of the feedbackcircuit stabilizing the dc. output voltage that is supplied to the loadof a wide range, where the transfer function feeding back the outputvoltage to the frequency of the carrier that drives the resonancecircuit is not provided with the pole located at the origin.

The patent reference 5 extends the application of the patent reference 3and claims the priority over the patent reference 3.

The patent reference 6 is the PCT application based on the patentreferences 3, 4, and 5.

The patent reference 3, the patent reference 5 and the patent reference6 are surveyed with relation to the present invention. These patentreferences are concerned with the stabilized dc. power supply generatingthe voltage with a resonance circuit.

Equivalent Power Supply Approximating DC. Power Supply

A power supply providing the stabilized dc. output voltage consists of avoltage generation circuit and the feedback circuit. The voltagegeneration circuit is composed of a driver circuit, the resonancecircuit, and the rectification circuit. The driver circuit generates ahigh-frequency alternating current carrier of a locally constantamplitude. The carrier drives the resonance circuit. The feedbackcircuit consists of an error amplifier and a voltage controlledoscillator (VCO). The error amplifier compares the dc. output voltagewith the reference voltage supplied externally to set the outputvoltage. The VCO is enabled to control the frequency of the carriergenerated by the driver circuit. The output voltage of the power supply,which is the output of the rectification circuit, is fed back to thefrequency of the carrier through the feedback circuit so as to bestabilized.

An equivalent power supply was developed in the patent references 3, 5and 6, where the equivalent power supply approximates the dc. stabilizedpower supply and can be analyzed by mathematical methods. The equivalentpower supply is schematically composed of a virtual voltage generationcircuit and a feedback circuit. The virtual voltage generation circuitincludes a driver circuit generating a carrier of locally fixedamplitude, a virtual resonance circuit driven by the carrier supplied bythe driver circuit, and a virtual rectification circuit generating dc.voltage from the output of the resonance circuit. The output of thepower supply is the output of the rectification circuit. The feedbackcircuit includes an error amplifier comparing the output voltage withthe reference voltage supplied externally to set the output voltage anda voltage controlled oscillator (VCO) generating the frequency decidedby the output of the error amplifier. The voltage controlled oscillatoris enabled to control the frequency of the carrier generated by thedriver circuit, and thus the output voltage is fed back to the frequencyof the carrier so as to be stabilized.

The virtual voltage generation circuit consists of a driver circuit, avirtual resonance circuit, and a virtual rectification circuit. Theresonance circuit converts the modulation of the carrier from thefrequency modulation at the input to the amplitude modulation at theoutput. The virtual resonance circuit, supplied with the carrier of thefrequency modulation as is the case with the resonance circuit, outputsthe envelop of the carrier modulated in amplitude, being different fromthe resonance circuit. The virtual rectification circuit inputs theenvelope, acts as a filter of first-order delay to the envelope, andoutputs an output equivalent to the output of the rectification circuit.

Operation of the equivalent power supply is described by a system ofdifferential equations, stability of which can be analyzedmathematically. The system of differential equations is derived and thenthe necessary conditions that the output voltage of the equivalent powersupply is stable in the neighborhood of the reference voltage are shown.An actual circuit which realizes stable feedback based on the necessaryconditions is shown with circuit constants given explicitly.

Frequency Modulation, Imaginary Resonance and Rectification Circuits

As for the resonance circuit whose transfer function given by h, lettingω_(r), Q, and g_(r), be the angular velocity of the resonance frequency,the Q-value, and the amplitude ratio at the resonance frequencyrespectively of the resonance, then δ, ω₀, and c are defined by

$\begin{matrix}{\delta = \frac{\omega_{r}}{2Q}} & \lbrack 1\rbrack \\{\omega_{0} = {\omega_{r}\sqrt{1 - \frac{1}{4Q^{2}}}}} & \lbrack 2\rbrack \\{c = \frac{g_{r}\omega_{r}}{Q}} & \lbrack 3\rbrack\end{matrix}$

and the resonance circuit is driven by such a carrier of fixed amplitudemodulated in frequency that is defined by

w exp(iω₀t+iψ)  [4]

where w is the amplitude of the carrier and ψ is a function of timerepresenting a shift of phase, then

$\begin{matrix}{\varphi = {\frac{}{t}\psi}} & \lbrack 5\rbrack\end{matrix}$

then the frequency of the carrier given in expression 4 is

ω₀+φ  [6]

and, letting r_(r) and r_(i) be defined by

$\begin{matrix}{r_{r} = {\frac{1}{2}c\; w}} & \lbrack 7\rbrack \\{r_{i} = {\frac{\delta}{2\omega_{0}}c\; w}} & \lbrack 8\rbrack\end{matrix}$

the resonance circuit, the transfer function of which is given inexpression 6, being driven by the carrier in expression 4, outputs thecarrier the amplitude of which is given by p and q as

√{square root over (p²+q²)}  [9]

where p and q satisfy

$\begin{matrix}{{\frac{}{t}p} = {{q\; \varphi} - {p\; \delta} + r_{r}}} & \lbrack 10\rbrack \\{{\frac{}{t}q} = {{{- p}\; \varphi} - {q\; \delta} + r_{i}}} & \lbrack 11\rbrack\end{matrix}$

Then a first order delay which, letting the dc. voltage generated byrectifying the carrier outputted by the resonance circuit be z, isrepresented by the following differential equation concerning z

$\begin{matrix}{{{\mu \frac{}{t}z} + z} = {\nu \sqrt{p^{2} + q^{2}}}} & \lbrack 12\rbrack\end{matrix}$

where μ and ν are a time constant and a multiplier at the rectificationcircuit respectively.

Feedback Circuit and System of Differential Equations

Letting k, d, E, A, and B be positive numbers and λ be a referencevoltage respectively, the dc. voltage z in expression 12 from therectification circuit is compared with the reference voltage λ and thevoltage difference between z and λ is fed back to the frequency of thecarrier φ in expression 5, where the feed back is expressed on theassumption that φ>0 by the transfer function having a pole located atthe origin as

$\begin{matrix}{\varphi = {k\; d\frac{\left( {E + {A\; s} + {B\; s^{2}}} \right)}{s}\left( {z - \lambda} \right)}} & \lbrack 13\rbrack\end{matrix}$

Then uniting expression 13, expressions 10, expression 11, andexpression 12 makes the system of differential equations describing thepower supply as

$\begin{matrix}{\mspace{20mu} {{\frac{}{t}p} = {{q\; \varphi} - {p\; \delta} + r_{r}}}} & \lbrack 14\rbrack \\{\mspace{20mu} {{\frac{}{t}q} = {{{- p}\; \varphi} - {q\; \delta} + r_{i}}}} & \lbrack 15\rbrack \\{\mspace{20mu} {{\frac{}{t}z} = \frac{{- z} + {\nu \sqrt{p^{2} + q^{2}}}}{\mu}}} & \lbrack 16\rbrack \\{{\frac{}{t}\varphi} = {{k\; E\; {d\left( {z - \lambda} \right)}} + \frac{k\; A\; {d\left( {{- z} + {\nu \sqrt{p^{2} + q^{2}}}} \right)}}{\mu} + {k\; B\; {d\left( {{- \frac{\sqrt{p^{2} + q^{2}}\left( {\nu + {\nu\mu\delta}} \right)}{\mu^{2}}} + \frac{z}{\mu^{2}} + \frac{\nu \left( {{q\; r_{i}} + {p\; r_{r}}} \right)}{\sqrt{p^{2} + q^{2}}\mu}} \right)}}}} & \lbrack 17\rbrack\end{matrix}$

Stability and Overshoots

The stability of the feedback in the stabilized dc. power supply isattributed to the stability of the system of the differential equationsgiven by expressions 14˜17. The stability of the system is decided bythe root of the characteristic polynomial derived from the system of thedifferential equations. That all the roots of the characteristicpolynomial have negative real parts is necessary and sufficientcondition that the system of differential equations is stable in thesense of Lyapunov. It is necessary condition to be satisfied by a powersupply that the system of differential equations describing the powersupply is stable in the sense of Lyapunov. Yet stability in the sense ofLyapunov is not enough for the stability of the power supply. Forinstance, there exists the cases where the power supply stable in thesense of Lyapunov oscillates the output voltage. There are also thecases where the output voltage oscillates in the neighborhood of thevoltage set by the reference voltage and the cases where the oscillationdecays in a long time

There is operation of the power supply which the system of differentialequations given by the expressions 14˜17 does not approximate. Theoutput voltage of the power supply is generated by the rectificationcircuit. The rectification circuit includes a capacitor, and the outputvoltage is buffered by the capacitor. As for the positive output voltageof the rectification circuit, pumping up the charge to the capacitorraises the output voltage, while it is impossible to pumping out thecharge from the capacitor so as to lower the output voltage. Then theraise of the output voltage can be described by the system of thedifferential equations 14˜17, while there are falls of the outputvoltage which the system of the differential equations can not reproducecorrectly.

While the output voltage is higher than the reference voltage, thefeedback works to lower the output voltage, reducing the currentsupplied to the rectification circuit by the resonance circuit. In thecase that the current is reduced to zero, while the output voltage ishigher than the reference voltage, the output voltage falls with a timeconstant of load resistance and the capacitor in the rectificationcircuit, where the fall is independent of the system of the differentialequations. When the output voltage becomes lower than the referencevoltage, the feedback begins to work, raising the output voltage. If theraise of the output voltage is accompanied by overshoots in voltage, theoutput voltage begins repeating the rise and the fall in voltage. Thenit is necessary for stable feedback that the rise of the output voltageis free from the overshoot.

A Sufficient Condition

In the case that the characteristic polynomial has the realcharacteristic root separated from imaginary roots, the raise of theoutput voltage is not accompanied by the overshoot and then the feedbackis stable. Let the transfer function feeding back the output voltage tothe frequency of the carrier be given by

$\begin{matrix}{\left( {\frac{E}{s} + 1 + {B\; s}} \right) \times N} & \lbrack 18\rbrack\end{matrix}$

where N is a positive constant, then 1/E is an approximation of the timeconstant where the output voltage is raised. So letting E and B beassigned as

$\begin{matrix}{E < \frac{1}{\mu}} & \lbrack 19\rbrack \\{and} & \; \\{B \sim \frac{1}{\delta}} & \lbrack 20\rbrack\end{matrix}$

the bandwidth for feedback be restricted by the time constant 1/E, N beassigned so that the loop gain become enough smaller than unity at thezero in the right half plane, then the feedback becomes stable

Zeros in Right Half Plane

A zero in the right half plane is created, for example, by the resonancecircuit driven by the carrier which is higher in frequency than theresonance frequency of the resonance circuit. In the stabilized dc.power supply where the output voltage is generated by rectifying theoutput of the resonance circuit and stabilized by feeding back theoutput voltage to the frequency of the carrier supplied to the resonancecircuit, in the case that a frequency range of the carrier is selectedto be higher than the resonance frequency of the resonance circuit asshown in FIG. 1, the output voltage is lowered by increasing and raisedby lowering the frequency of the carrier. Yet the fall of the frequencycauses drop of the output voltage. Namely, to lower the frequency inorder to increase the output voltage causes temporary drop of the outputvoltage.

When the frequency of the carrier is lowered and moved to the resonanceto raise the output voltage, the fall of the frequency causes immediatedrop of the output voltage. The amplitude of the carrier outputted bythe resonance circuit is changed after the frequency is shifted. Thetime delay from the shift of the frequency to the change of theamplitude is approximated by 1/δ. The amplitude of the temporary drop isdependent of a magnitude of the load. That the control which makes theoutput voltage increase accompanies the temporal drop of the outputvoltage is characteristic of the control system provided with zeroslocated at the right half plane.

DISCLOSURE OF INVENTION Technical Problem

In the stabilized dc. voltage power supply where the output voltagegenerated by rectifying the output of the resonance circuit isstabilized by frequency modulation of the carrier supplied to theresonance circuit, the feedback circuit and its constants, improving thefrequency response of the output voltage limited by a large delay causedby the feedback of the output voltage to the frequency of the carrierare given.

Technical Solution

Simultaneously implementing feedback of a large delay returning theoutput voltage to the frequency of the carrier and feedback of a smalldelay returning the output voltage to the amplitude of the carrierrepresses rapid frequency change of the carrier and improves thefrequency response of the output voltage.

Variable Amplitude of Carrier

Letting w+x be the amplitude of the carrier where w is a constant and xis a function of time and ψ be a function of time, the carrier modulatedin frequency is represented by

(w+x)exp(i ω₀t+iψ)  [21]

and φ is defined by ψ in expression 21 as

$\begin{matrix}{\varphi = {\frac{}{t}\psi}} & \lbrack 22\rbrack\end{matrix}$

then the frequency of the carrier in expression 21 with frequencymodulation is given by

ω₀+φ  [23]

Then r_(r) and r_(i) are modified from the definitions in expressions 7and 8, and redefined by

$\begin{matrix}{r_{r} = {\frac{1}{2}c}} & \lbrack 24\rbrack \\{r_{i} = {\frac{\delta}{2\omega_{0}}c}} & \lbrack 25\rbrack\end{matrix}$

The amplitude of the carrier supplied by the resonance circuit driven bythe carrier shown in expression 21 is given by solutions of thefollowing simultaneous equations

$\begin{matrix}{{\frac{}{t}p} = {{q\; \varphi} - {p\; \delta} + {r_{r}\left( {w + x} \right)}}} & \lbrack 26\rbrack \\{{\frac{}{t}q} = {{{- p}\; \varphi} - {q\; \delta} + {r_{i}\left( {w + x} \right)}}} & \lbrack 27\rbrack \\{as} & \; \\\sqrt{p^{2} + q^{2}} & \lbrack 28\rbrack\end{matrix}$

Let z be a dc. voltage obtained by rectifying and smoothing the carriersupplied by the resonance circuit, then z satisfies the followingequation

$\begin{matrix}{{{\mu \frac{}{t}z} + z} = {\nu \sqrt{p^{2} + q^{2}}}} & \lbrack 29\rbrack\end{matrix}$

where μ is a time constant at smoothing and ν is a multiplier at therectification.

Feedback of Output Voltage to Amplitude

Letting an error be voltage difference between z and λ, where z is theoutput of the rectification circuit and λ is the reference voltagesetting up the output voltage of the power supply, the error is fed backto amplitude w+x of the carrier by

x=−G(z−λ)  [30]

where G is a positive constant. As is shown by expression 30, z, beinghigher than the reference voltage, makes x negative and works to lowerthe output voltage z. Similarly z higher than the reference voltageworks to raise the output voltage.

Substituting expression 30 for the amplitude in expressions 26 and 27gives

$\begin{matrix}{{\frac{}{t}p} = {{\varphi \; q} - {\delta \; p} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{r}}}} & \lbrack 31\rbrack \\{{\frac{}{t}q} = {{{- \varphi}\; p} - {\delta \; q} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{i}}}} & \lbrack 32\rbrack\end{matrix}$

Feedback of Output Voltage to Frequency of Carrier

Letting an error be voltage difference between z and λ, where z is theoutput of the rectification circuit and λ is the reference voltagesetting up the output voltage of the power supply, the transfer functionfeeding back the error to the frequency of the carrier is shown withpositive constants k, d, E, A, B and under the condition that φ≧0 by

$\begin{matrix}{\varphi = {{kd}\frac{\left( {E + {As} + {Bs}^{2}} \right)}{s}\left( {z - \lambda} \right)}} & \lbrack 33\rbrack\end{matrix}$

where φ is defined in expression 22 and z is defined in expression 29.

Second Derivative of Output Voltage

Derivative of the output voltage z is given in expression 29, from whichthe second derivative of z relative to time t is obtained as follows.

$\begin{matrix}{{\frac{^{2}}{t^{2}}z} = {\frac{1}{\mu}\left( {{{- \frac{}{t}}z} + {\frac{1}{2}\frac{v\left( {{2\; p\frac{}{t}p} + {2\; q\frac{}{t}q}} \right)}{\sqrt{p^{2} + q^{2}}}}} \right)}} & \lbrack 34\rbrack\end{matrix}$

Expressions 29, 31 and 32 transform expression 34 to

$\begin{matrix}{{\frac{^{2}}{t^{2}}z} = {\frac{{v\left( {w + x} \right)}\left( {{qr}_{i} + {r_{r}p}} \right)}{\mu \sqrt{p^{2} + q^{2}}} - \frac{{v\sqrt{p^{2} + q^{2}}} + {\sqrt{p^{2} + q^{2}}v\; {\mu\delta}} - z}{\mu^{2}}}} & \lbrack 35\rbrack\end{matrix}$

Expression 35 is reduced with expression 30 to

$\begin{matrix}{{\frac{^{2}}{t^{2}}z} = {\frac{{v\left( {w - {G\left( {z - \lambda} \right)}} \right)}\left( {{qr}_{i} + {r_{r}p}} \right)}{\mu \sqrt{p^{2} + q^{2}}} + \frac{{{- v}\sqrt{p^{2} + q^{2}}} - {\sqrt{p^{2} + q^{2}}v\; {\mu\delta}} + z}{\mu^{2}}}} & \lbrack 36\rbrack\end{matrix}$

Transfer Function From Output Voltage to Frequency of Carrier

Expression 33 is rewritten to

$\begin{matrix}{{\frac{}{t}\varphi} = {{kd}\left( {{E\left( {z - \lambda} \right)} + {A\frac{}{t}z} + {B\frac{^{2}}{t^{2}}z}} \right)}} & \lbrack 37\rbrack\end{matrix}$

Substituting expression 29 for the first derivative of z and expression36 for the second derivative gives the following expression.

$\begin{matrix}{{\frac{}{t}\varphi} = {{kd} {\quad\left( {{E\left( {z - \lambda} \right)} + \frac{A\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu} + {B\left( {\frac{{v\left( {w - {G\left( {z - \lambda} \right)}} \right)}\left( {{qr}_{i} + {r_{r}p}} \right)}{\mu \sqrt{p^{2} + q^{2}}} + \frac{{{- v}\sqrt{p^{2} + q^{2}}} - {\sqrt{p^{2} + q^{2}}v\; {\mu\delta}} + z}{\mu^{2}}} \right)}} \right)}}} & \lbrack 38\rbrack\end{matrix}$

Description of Power Supply by Differential Equations

A regular system of differential equations composed of expressions 38,31, 32, 29 describes the power supply where output voltage is fed backto the frequency and the amplitude of the carrier.

$\begin{matrix}{\mspace{79mu} {{\frac{}{t}p} = {{\varphi \; q} - {\delta \; p} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{r}}}}} & \lbrack 39\rbrack \\{\mspace{79mu} {{\frac{}{t}q} = {{{- \varphi}\; p} - {\delta \; q} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{i}}}}} & \lbrack 40\rbrack \\{\mspace{79mu} {{\frac{}{t}z} = \frac{{- z} + {v\sqrt{p^{2} + q^{2}}}}{\mu}}} & \lbrack 41\rbrack \\{{\frac{}{t}\varphi} = {{kd} {\quad\left( {{E\left( {z - \lambda} \right)} + \frac{A\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu} + {B\left( {\frac{{v\left( {w - {G\left( {z - \lambda} \right)}} \right)}\left( {{qr}_{i} + {r_{r}p}} \right)}{\mu \sqrt{p^{2} + q^{2}}} + \frac{{{- v}\sqrt{p^{2} + q^{2}}} - {\sqrt{p^{2} + q^{2}}v\; {\mu\delta}} + z}{\mu^{2}}} \right)}} \right)}}} & \lbrack 42\rbrack\end{matrix}$

Equilibrium point of Differential Equation System

An equilibrium point of the regular system of the differential equationsis defined by

$\begin{matrix}{{{\frac{}{t}p} = {{0.\frac{}{t}q} = 0}},{{\frac{}{t}z} = 0},{{\frac{}{t}\varphi} = 0}} & \lbrack 43\rbrack\end{matrix}$

The equilibrium point satisfies expressions 34 and 43

$\begin{matrix}{{\frac{^{2}}{t^{2}}z} = 0} & \lbrack 44\rbrack\end{matrix}$

The equilibrium point is the simultaneous solutions of the followinglinear equations.

$\begin{matrix}{\mspace{79mu} {0 = {{\varphi \; q} - {\delta \; p} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{r}}}}} & \lbrack 45\rbrack \\{\mspace{79mu} {0 = {{{- \varphi}\; p} - {\delta \; q} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{i}}}}} & \lbrack 46\rbrack \\{\mspace{79mu} {0 = \frac{{- z} + {v\sqrt{p^{2} + q^{2}}}}{\mu}}} & \lbrack 47\rbrack \\{0 = {{kd} {\quad\left( {{E\left( {z - \lambda} \right)} + \frac{A\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu} + {B\left( {\frac{{v\left( {w - {G\left( {z - \lambda} \right)}} \right)}\left( {{qr}_{i} + {r_{r}p}} \right)}{\mu \sqrt{p^{2} + q^{2}}} + \frac{{{- v}\sqrt{p^{2} + q^{2}}} - {\sqrt{p^{2} + q^{2}}v\; {\mu\delta}} + z}{\mu^{2}}} \right)}} \right)}}} & \lbrack 48\rbrack\end{matrix}$

Letting the equilibrium point be denoted by

p_(e), q_(e), z_(e), φ_(e), expressions 37 and 44 leads to

z_(e)=λ  [49]

It can be seen from expression 30 that x=0 at the equilibrium point. Foran arbitrary positive w which is enough to generate the output voltagez_(e), there exists such the equilibrium point that frequency φ_(e)generates z_(e) with the amplitude w. Let r be denoted by

r=√{square root over (φ_(e) ²+δ²)}  [50]

then p_(e), q_(e), z_(e) and λ are given as functions of φ_(e) by

$\begin{matrix}{p_{e} = \frac{w\left( {{\delta \; r_{r}} + {r_{i}\varphi_{e}}} \right)}{\varphi_{e}^{2} + \delta^{2}}} & \lbrack 51\rbrack \\{q_{e} = {- \frac{w\left( {{{- \delta}\; r_{i}} + {\varphi_{e}r_{r}}} \right)}{\varphi_{e}^{2} + \delta^{2}}}} & \lbrack 52\rbrack \\{z_{e} = \frac{vwr}{\sqrt{\varphi_{e}^{2} + \delta^{2}}}} & \lbrack 53\rbrack \\{\lambda = \frac{vwr}{\sqrt{\varphi_{e}^{2} + \delta^{2}}}} & \lbrack 54\rbrack\end{matrix}$

Stability of Differential Equation System

Stability of the differential equation system is analyzed by theLyapunov method in the neighborhood of the equilibrium point. For thepurpose, p, q, z and φ are expanded in the neighborhood of p_(e), q_(e),z_(e) and φ_(e) as follows.

p=p _(e) +Δp  [55]

q=q _(e) +Δq  [56]

z=z _(e) +Δz  [57]

φ=φ_(e)+Δφ  [58]

Substituting the above equations for expressions 39˜42 and ignoring thehigher order terms of the expansion give the following regular system ofthe differential equations relative to Δp, Δq, Δz and Δφ.

$\begin{matrix}{\mspace{79mu} (1)} & \; \\{\mspace{79mu} {{\frac{}{t}\begin{bmatrix}{\Delta \; p} \\{\Delta \; q} \\{\Delta \; z} \\{\Delta\varphi}\end{bmatrix}} = {M\begin{bmatrix}{\Delta \; p} \\{\Delta \; q} \\{\Delta \; z} \\{\Delta\varphi}\end{bmatrix}}}} & \lbrack 59\rbrack \\{\mspace{79mu} (2)} & \; \\{\mspace{79mu} {{M = \begin{bmatrix}{- \delta} & \varphi_{e} & {- {Gr}_{r}} & q_{e} \\{- \varphi_{e}} & {- \delta} & {- {Gr}_{i}} & {- p_{e}} \\\frac{{vp}_{e}}{\sqrt{p_{e}^{2} + {q_{e}^{2}\mu}}} & \frac{{vq}_{e}}{\sqrt{p_{e}^{2} + {q_{e}^{2}\mu}}} & {- \mu^{- 1}} & 0 \\M_{41} & M_{42} & M_{43} & 0\end{bmatrix}}\mspace{79mu} {where}}} & \lbrack 60\rbrack \\{M_{41} = {\frac{kdv}{{\mu^{2}\left( {p_{e}^{2} + q_{e}^{2}} \right)}^{3/2}}\left( {{q_{e}B\; {\mu \left( {w - {z_{e}G} + {\lambda \; G}} \right)}\left( {{{- r_{i}}p_{e}} + {q_{e}r_{r}}} \right)} - {{p_{e}\left( {p_{e}^{2} + q_{e}^{2}} \right)}\left( {{{- A}\; \mu} + B + {B\; {\mu\delta}}} \right)}} \right)}} & \lbrack 61\rbrack \\{M_{42} = {\frac{kdv}{{\mu^{2}\left( {p_{e}^{2} + q_{e}^{2}} \right)}^{3/2}}\left( {{{- p_{e}}B\; {\mu \left( {w - {zG} + {\lambda \; G}} \right)}\left( {{{- r_{i}}p_{e}} + {q_{e}r_{r}}} \right)} - {{q_{e}\left( {p_{e}^{2} + q_{e}^{2}} \right)}\left( {{{- A}\; \mu} + B + {B\; {\mu\delta}}} \right)}} \right)}} & \lbrack 62\rbrack \\{M_{43} = {{kd}\left( {E - \frac{A}{\mu} - \frac{B\left( {{- \sqrt{p_{e}^{2} + q_{e}^{2}}} + {v\; \mu \; p_{e}r_{r}G} + {v\; \mu \; q_{e}r_{i}G}} \right)}{\mu^{2}\sqrt{p_{e}^{2} + q_{e}^{2}}}} \right)}} & \lbrack 63\rbrack\end{matrix}$

Elements of matrix M are functions of p_(e), q_(e), z_(e) and φ_(e),while considering expressions 51˜54, the elements are thought to befunctions of φ_(e). Then let the characteristic polynomial of M be m(h)and m(h) be represented by

m(h)=a ₀ h ⁴ +a ₁ h ³ +a ₂ h ² +a ₃ h+a ₄  [64]

then coefficients a₀, a₁, a₂, a₃ and a₄ are given as functions of φ_(e)in the following.

$\begin{matrix}{a_{0} = 1} & \lbrack 65\rbrack \\{a_{1} = \frac{{2{\mu\delta}} + 1}{\mu}} & \lbrack 66\rbrack \\{a_{2} = \frac{{\varphi_{e}{kdvBwr}} + {{vG}\; \delta \; r} + {2\delta \sqrt{\varphi_{e}^{2} + \delta^{2}}} + {\left( {\varphi_{e}^{2} + \delta^{2}} \right)^{3/2}\mu}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 67\rbrack \\{a_{3} = \frac{{\varphi_{e}{kdvwAr}} + \left( {\varphi_{e}^{2} + \delta^{2}} \right)^{3/2} + {{vGr}\; \left( {\varphi_{e}^{2} + \delta^{2}} \right)}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 68\rbrack \\{a_{4} = \frac{\varphi_{e}{kdvwEr}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 69\rbrack\end{matrix}$

where r is defined by

r=√{square root over (r_(r) ² +r _(i) ²)}  [70]

Let feeding back the output voltage to the amplitude be given not byexpression 30 but by

$\begin{matrix}{x = {{- {G\left( {z - \lambda} \right)}} - {H\frac{z}{t}}}} & \lbrack 71\rbrack\end{matrix}$

then coefficients a₀, a₁, a₂, a₃ and a₄ are as follows.

$\begin{matrix}{a_{0} = 1} & \lbrack 72\rbrack \\{a_{1} = \frac{{\delta \; {vHr}} + {2{\delta\mu}\sqrt{\varphi_{e}^{2} + \delta^{2}}} + \sqrt{\varphi_{e}^{2} + \delta^{2}}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 73\rbrack \\{a_{2} = \frac{\begin{matrix}{{2\delta \sqrt{\varphi_{e}^{2} + \delta^{2}}} + {\delta \; {vrG}} +} \\{{\varphi_{e}{kdvwrB}} + {\mu \left( {\varphi_{e}^{2} + \delta^{2}} \right)}^{3/2} + {{rHv}\left( {\varphi_{e}^{2} + \delta^{2}} \right)}}\end{matrix}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 74\rbrack \\{a_{3} = \frac{{\varphi_{e}{wrkdvA}} + \left( {\varphi_{e}^{2} + \delta^{2}} \right)^{3/2} + {{vrG}\left( {\varphi_{e}^{2} + \delta^{2}} \right)}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 75\rbrack \\{a_{4} = \frac{\varphi_{e}{kdvwrE}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 76\rbrack\end{matrix}$

Characteristic polynomial m(h) is a function of h and φ_(e), φ can besubstituted without confusion for φ_(e) in the expression of thecharacteristic polynomials.

Characteristic Polynomial For Fixed Carrier Frequency

In the case that the frequency of the carrier is fixed and that theoutput voltage is not fed back to the frequency of the carrier, theoutput voltage is stabilized by feeding back the output voltage to theamplitude of the carrier. Let the output voltage be fed back to theamplitude of the carrier by

x=−G(z−λ)  [77]

where G is a positive constant, the simplest differential equationsystem describing the power supply will be

$\begin{matrix}{{\frac{}{t}p} = {{\varphi \; q} - {\delta \; p} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{r}}}} & \lbrack 78\rbrack \\{{\frac{}{t}q} = {{{- \varphi}\; p} - {\delta \; q} + {\left( {w - {G\left( {z - \lambda} \right)}} \right)r_{i}}}} & \lbrack 79\rbrack \\{{\frac{}{t}z} = \frac{{- z} + {v\sqrt{p^{2} + q^{2}}}}{\mu}} & \lbrack 80\rbrack\end{matrix}$

In expressions 78˜80, φ is fixed. So p_(e), q_(e) and z_(e) areexpressed by the fixed φ and λ prescribed beforehand with

$\begin{matrix}{{r = \sqrt{\varphi^{2} + \delta^{2}}}{as}} & \lbrack 81\rbrack \\{p_{e} = \frac{{\delta \; r_{r}w} + {\varphi \; r_{i}w} + {\varphi \; r_{i}G\; \lambda} + {\delta \; r_{r}G\; \lambda}}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\sqrt{\varphi^{2} + \delta^{2}} + {v\; {Gr}}} \right)}} & \lbrack 82\rbrack \\{q_{e} = \frac{{\delta \; r_{i}w} + {\delta \; r_{i}G\; \lambda} - {r_{r}w\; \varphi} - {r_{r}G\; \varphi \; \lambda}}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\sqrt{\varphi^{2} + \delta^{2}} + {vGr}} \right)}} & \lbrack 83\rbrack \\{z_{e} = \frac{{vr}\left( {w + {\lambda \; G}} \right)}{\sqrt{\varphi^{2} + \delta^{2}} + {vGr}}} & \lbrack 84\rbrack\end{matrix}$

Let m(h) be the characteristic polynomial of the differential equationsystem derived from 78˜80 and describing deviation from the equilibriumpoint and m(h) be represented by

m(h)=a ₀ h ³ +a ₁ h ² +a ₂ h ¹ +a ₃  [85]

coefficients a₀, a₁, a₂ and a₃ are given by

$\begin{matrix}{a_{0} = 1} & \lbrack 86\rbrack \\{a_{1} = \frac{1 + {2\delta \; \mu}}{\mu}} & \lbrack 87\rbrack \\{a_{2} = \frac{{r\; \delta \; {vG}} + {\mu \left( {\varphi^{2} + \delta^{2}} \right)}^{3/2} + {2\delta \sqrt{\varphi^{2} + \delta^{2}}}}{\sqrt{\varphi^{2} + \delta^{2}}\mu}} & \lbrack 88\rbrack \\{a_{3} = \frac{\sqrt{\varphi^{2} + \delta^{2}}\left( {\sqrt{\varphi^{2} + \delta^{2}} + {Gvr}} \right)}{\mu}} & \lbrack 89\rbrack\end{matrix}$

Characteristic polynomial m(h) is resolved as

$\begin{matrix}{{m(h)} = {\frac{\left( {h^{2} + \varphi^{2} + \delta^{2} + {2h\; \delta}} \right)\left( {{h\; \mu} + 1} \right)}{\mu} + \frac{{{vr}\left( {{h\; \delta} + \varphi^{2} + \delta^{2}} \right)}G}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}} & \lbrack 90\rbrack\end{matrix}$

Voltage difference between the output voltage z and the referencevoltage λ is fed back to the amplitude of the carrier. Let contributionto the output voltage made by feeding back υ to the amplitude berepresented by

gυ  [91]

then the output voltage is given by

z=gυ  [92]

In the other hand, at the subtraction circuit,

υ=λ−z  [93]

then

$\begin{matrix}{z = {\frac{g\; \lambda}{1 + g}.}} & \lbrack 94\rbrack\end{matrix}$

Therefore

0=1+g  [95]

and the following characteristic polynomial

0=m(h)  [96]

are equivalent conditions for stability, and expression 95 and 96 areequivalent. Then

$\begin{matrix}{g = \frac{{vGr}\left( {{h\; \delta} + \varphi^{2} + \delta^{2}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {h^{2} + \varphi^{2} + \delta^{2} + {2h\; \delta}} \right)\left( {{h\; \mu} + 1} \right)}} & \lbrack 97\rbrack\end{matrix}$

In the case that the feedback to the amplitude is given not byexpression 77, but by the following expression

$\begin{matrix}{x = {{- {G\left( {z - \lambda} \right)}} - {H\frac{}{t}z}}} & \lbrack 98\rbrack\end{matrix}$

where G and H are positive constants. Let m_(B)(h) be the characteristicpolynomial m(h) corresponding to expression 90, m_(B)(h) is given by

$\begin{matrix}{{m_{B}(h)} = {\frac{\left( {{2h\; \delta} + h^{2} + \delta^{2} + \varphi^{2}} \right)\left( {{h\; \mu} + 1} \right)}{\mu} + \frac{{{vr}\left( {{h\; \delta} + \varphi^{2} + \delta^{2}} \right)}\left( {{hH} + G} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}} & \lbrack 99\rbrack\end{matrix}$

and expression 97 is rewritten to

$\begin{matrix}{g = \frac{{{vr}\left( {{h\; \delta} + \varphi^{2} + \delta^{2}} \right)}\left( {{hH} + G} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {{2h\; \delta} + h^{2} + \delta^{2} + \varphi^{2}} \right)\left( {{h\; \mu} + 1} \right)}} & \lbrack 100\rbrack\end{matrix}$

correspondingly.Resolution of Characteristic Polynomial m(h)

Polynomial m(h), which is the characteristic polynomial with thecoefficients given in expressions 65˜69, is represented by

$\begin{matrix}{{m(h)} = {\frac{{h\left( {\varphi^{2} + h^{2} + {2h\; \delta} + \delta^{2}} \right)}\left( {{\mu \; h} + 1} \right)}{\mu} + \frac{w\; \varphi \; {k\left( {E + {hA} + {h^{2}B}} \right)}{vrd}}{\sqrt{\varphi^{2} + \delta^{2}}\mu} + \frac{{vrhG}\left( {\varphi^{2} + {h\; \delta} + \delta^{2}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}} & \lbrack 101\rbrack\end{matrix}$

The second term on the right hand side in expression 101 coincides withthe numerator of the transfer function for the feedback. Therefore thesecond term originates in the feedback circuit. The first and the thirdterms independent of the feedback stem from the voltage generationcircuit. As for the first term, h corresponds to the pole located at theorigin created by the feedback circuit, and (μh+1) to the rectificationcircuit whose time constant is 1/μ.

The subtraction circuit in the error amplifier outputs voltagedifference υ between the output voltage z and the reference voltage. Thevoltage difference υ is fed back to the frequency and the amplitude ofthe carrier. Let the contribution of the output voltage made by feedingback υ to the frequency be represented by

fυ  [102]

and the contribution of the output voltage made by feeding back υ to theamplitude be expressed by

gυ  [103]

then the output voltage z is given by

z=fυ+gυ  [104]

In the other hand, at the subtraction circuit,

υ=λ−z  [105]

then z is represented by

$\begin{matrix}{z = \frac{\left( {f + g} \right)\lambda}{1 + f + g}} & \lbrack 106\rbrack\end{matrix}$

The following equations

0=1+f+g  [107]

and

0=m(h)  [108]

where m(h) is the characteristic polynomial. Hence f and g are givenexcept for a common multiplier by

$\begin{matrix}{f = \frac{k\; \varphi \; {w\left( {E + {Ah} + {Bh}^{2}} \right)}}{\sqrt{\varphi^{2} + \delta^{2}}{h\left( {\varphi^{2} + h^{2} + \delta^{2} + {2h\; \delta}} \right)}\left( {1 + {h\; \mu}} \right)}} & \lbrack 109\rbrack \\{g = \frac{{Gvr}\left( {\varphi^{2} + {h\; \delta} + \delta^{2}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + h^{2} + \delta^{2} + {2h\; \delta}} \right)\left( {1 + {h\; \mu}} \right)}} & \lbrack 110\rbrack\end{matrix}$

The characteristic polynomial m(h) with coefficients a₀, a₁, a₂, a₃ anda₄ given in expressions 72˜76 is represented by

$\begin{matrix}{{m(h)} = {\frac{{h\left( {\delta^{2} + {2h\; \delta} + h^{2} + \varphi^{2}} \right)}\left( {{h\; \mu} + 1} \right)}{\mu} + \frac{\varphi \; {{wk}\left( {E + {hA} + {h^{2}B}} \right)}{vrd}}{\sqrt{\varphi^{2} + \delta^{2}}\mu} + \frac{{{hvr}\left( {\delta^{2} + {h\; \delta} + \varphi^{2}} \right)}\left( {G + {hH}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}} & \lbrack 111\rbrack\end{matrix}$

For the above mentioned m(h), f in expression 109 and g in expression110 are given as

$\begin{matrix}{f = \frac{k\; \varphi \; {w\left( {E + {Ah} + {Bh}^{2}} \right)}}{\sqrt{\varphi^{2} + \delta^{2}}{h\left( {\varphi^{2} + h^{2} + \delta^{2} + {2h\; \delta}} \right)}\left( {1 + {h\; \mu}} \right)}} & \lbrack 112\rbrack \\{g = \frac{{{vr}\left( {\varphi^{2} + {h\; \delta} + \delta^{2}} \right)}\left( {G + {hH}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + h^{2} + \delta^{2} + {2h\; \delta}} \right)\left( {1 + {h\; \mu}} \right)}} & \lbrack 113\rbrack\end{matrix}$

As for f, let a real root of the second order equation

E+Ah+Bh ²=0  [114]

in the numerator cancel 1+μh in the denominator, then the numerator of fis of the first degree in h and the denominator is of the third degreein h where one of the pole is located at the origin. As for g, let G+Hhin the numerator cancel 1+μh in the denominator, then the numerator of gis of the first degree in h, and the denominator is of the second degreein h.

Transfer Function From Amplitude to Output Voltage

Let the output voltage be fed back to the amplitude by

x=−G(z−λ)  [115]

for a positive constant G, then g is given by expression 110, while by

$\begin{matrix}{x = {{- {D\left( {z - \lambda} \right)}} - {H\frac{}{t}z}}} & \lbrack 116\rbrack\end{matrix}$

for positive constants G and H, then g is given by

$\begin{matrix}{g = \frac{{{vr}\left( {\varphi^{2} + \delta^{2} + {h\; \delta}} \right)}\mspace{11mu} \left( {{Hh} + G} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + \delta^{2} + h^{2} + {2\; h\; \delta}} \right)\mspace{11mu} \left( {{h\; \mu} + 1} \right)}} & \lbrack 117\rbrack\end{matrix}$

From expressions 110 and 117, the transfer function from the amplitudeto the output voltage are found to be

$\begin{matrix}\frac{{vr}\left( {\varphi^{2} + \delta^{2} + {h\; \delta}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + \delta^{2} + h^{2} + {2\mspace{11mu} h\; \delta}} \right)\mspace{11mu} \left( {{h\; \mu} + 1} \right)} & \lbrack 118\rbrack\end{matrix}$

Since the transfer function includes the first order delay

$\begin{matrix}\frac{1}{{\mu \; h} + 1} & \lbrack 119\rbrack\end{matrix}$

due to the rectification circuit, The voltage difference z−λ transformedso as to cancel the delay

μh+1,

before applying the amplitude, can be fed back to That amplitude faster.

Compensated Transfer Function From Voltage Difference to Output Voltage

In the case that the delay in the rectification circuit, the transferfunction from the voltage difference to the output voltage iscompensated is given by

$\begin{matrix}\frac{{vr}\left( {\varphi^{2} + {h\; \delta} + \delta^{2}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + {2\mspace{11mu} h\; \delta} + \delta^{2} + h^{2}} \right)} & \lbrack 121\rbrack\end{matrix}$

Then the dc. gain at h=0 is expressed as

$\begin{matrix}{\frac{vr}{\sqrt{\varphi^{2} + \delta^{2}}}.} & \lbrack 122\rbrack\end{matrix}$

Namely, letting w be the amplitude of the carrier, the output voltage is

$\begin{matrix}{\frac{vrw}{\sqrt{\varphi^{2} + \delta^{2}}},} & \lbrack 123\rbrack\end{matrix}$

coinciding with expression 54.

The expression 121 shows the delay that the voltage difference comes outto the output voltage when the voltage difference is applied to theamplitude

Schematic Diagram for Model of Feedback

Let the voltage difference between the output voltage and the referencevoltage, λ−z, be supplied to the circuits A and B as shown in FIG. 2,and the output voltage z be the sum of the output of the individualcircuits as

A(λ−z)+B(λ−z)  [124]

where the transfer functions of circuit A and B are denoted by A and Brespectively. Then z is given by

$\begin{matrix}{z = \frac{\lambda \left( {A + B} \right)}{1 + A + B}} & \lbrack 125\rbrack\end{matrix}$

So the stability of the feedback is attributed to the roots of thefollowing equation:

1+A+B=0.  [126]

For simplicity, let the transfer function A be

$\begin{matrix}{A = \frac{a}{s}} & \lbrack 127\rbrack\end{matrix}$

with the pole at the origin, and B be the simplest model of an amplifiersuch as

$\begin{matrix}{B = \frac{c}{s + b}} & \lbrack 128\rbrack\end{matrix}$

then expression 126 is rewritten to

s ² +s(a+b+c)+ab=0.  [129]

Necessary and sufficient conditions that the expression 129 has realnegative roots is

$\begin{matrix}{{\frac{\left( {a + b + c} \right)^{2}}{4} - {ab}} \geq 0} & \lbrack 130\rbrack\end{matrix}$

and expression 130 holds for arbitrary positive a, b and c.

Let roots of expression 129 be α and β, where α is closest to theorigin. Expression 129 is rewritten as

s ² +s(b+c)=−a(s+b)  [131]

Then as −a grows in magnitude, −α and −β become large, and it can beseen that

(b+c)²−(b+c)(a+b+c)+ab<0,  [132]

therefore

β<−(b+c)  [133]

The loop including circuit A feeds back the output voltage to thefrequency of the carrier, and the loop including circuit B feeds backthe output voltage to the amplitude of the carrier. Since circuit B iswider than circuit A in bandwidth and higher in gain, the output voltageis mainly fed back to the amplitude. The transfer function of circuit Bis not provided with the pole at the origin, and then there is thestandard deviation between the output of circuit B and the referencevoltage. The standard deviation is removed by the output of circuit A.

Transfer Function Compensating Delay of Rectification Circuit

Let μ₁ be such that

μ₁˜μ  [134]

and the output voltage is fed back to the amplitude of the carrier by

$\begin{matrix}{x = {{- {G\left( {z - \lambda} \right)}} - {\mu_{1}G\frac{}{t}z}}} & \lbrack 135\rbrack\end{matrix}$

where G is a positive constant. Then g is given by

$\begin{matrix}{g = {\frac{{{vr}\left( {\varphi^{2} + \delta^{2} + {h\; \delta}} \right)}\mspace{11mu} G\; \left( {{\mu_{1}h} + 1} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + \delta^{2} + h^{2} + {2\mspace{11mu} h\; \delta}} \right)\mspace{11mu} \left( {{h\; \mu} + 1} \right)}.}} & \lbrack 136\rbrack\end{matrix}$

Similarly f is written by

$\begin{matrix}{f = {\frac{k\; \varphi \; {w\left( {E + {A\; h} + {Bh}^{2}} \right)}}{\sqrt{\varphi^{2} + \delta^{2}}{h\left( {\varphi^{2} + h^{2} + \delta^{2} + {2\mspace{11mu} h\; \delta}} \right)}\mspace{11mu} \left( {1 + {h\; \mu}} \right)}.}} & \lbrack 137\rbrack\end{matrix}$

As for E+Ah+Bh² in the numerator of f, it is expected that one of thereal roots of the following equation

E+Ah+Bh ²=0  [138]

cancels 1+μh in the denominator. Namely expression 138 has a real rootin the neighborhood of −1/μ.Roots of Equation m(h)=0

In order to examine the roots of equation m(h)=0 derived from thecharacteristic polynomial, let f₁(h) and f₂(h) be defined as

$\begin{matrix}{{f_{1}(h)} = {\frac{{h\left( {\varphi^{2} + h^{2} + {2\mspace{11mu} h\; \delta} + \delta^{2}} \right)}\mspace{11mu} \left( {{h\; \mu} + 1} \right)}{\mu} + \frac{{{hvr}\left( {\varphi^{2} + {h\; \delta} + \delta^{2}} \right)}\mspace{11mu} \left( {G + {hH}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}} & \lbrack 139\rbrack \\{\mspace{79mu} {{{f_{2}(h)} = {- \frac{k\; w\; {\varphi \left( {{hA} + {h^{2}B} + E} \right)}{vrd}}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}}\mspace{79mu} {Then}}} & \lbrack 140\rbrack \\{\mspace{79mu} {{m(h)} = {{f_{1}(h)} - {f_{2}(h)}}}} & \lbrack 141\rbrack\end{matrix}$

Let m(h) be represented by

m(h)=a ₀ h ⁴ +a ₁ h ³ +a ₂ h ² +a ₃ h+a ₄,  [142]

then coefficients a₀ and a₁ are

$\begin{matrix}{{a_{0} = 1}{and}} & \lbrack 143\rbrack \\{a_{1} = {\frac{\delta \; {vG}\; \mu_{1}r}{\sqrt{\varphi^{2} + \delta^{2}}\mu} + {\frac{{2\; {\delta\mu}} + 1}{\mu}.}}} & \lbrack 144\rbrack\end{matrix}$

So μ₁ is approximate to μ, and

$\begin{matrix}\frac{\nu \; w\; r}{\sqrt{\varphi_{e}^{2} + \delta^{2}}} & \lbrack 145\rbrack\end{matrix}$

is equal to the reference voltage λ from expression 54 and to the outputvoltage, letting G be selected so as to satisfy

$\begin{matrix}{{\frac{G}{w}1},} & \lbrack 146\rbrack\end{matrix}$

then, letting the sum of the root of m(h)=0 be S, S is evaluated as

$\begin{matrix}{{- S} = {a_{1} \geq {{2\delta} + \frac{1}{\mu}}}} & \lbrack 147\rbrack\end{matrix}$

Polynomial f₁(h) is the characteristic polynomial where the outputvoltage is fed back to the frequency of the carrier. Letting α be theroot between 0 and −1/μ, and β be the root next to α, the feedback isstabilized, for example, by such arrangement of the roots as

$\begin{matrix}{{- \frac{1}{2\delta}} < \beta < {- \frac{1}{\mu}} < \alpha < 0} & \lbrack 148\rbrack\end{matrix}$

where α and β are roughly given by the term E+as+bs² as

$\begin{matrix}{\alpha \sim {- \frac{E}{A}}} & \lbrack 149\rbrack \\{\beta \sim {- \frac{A}{2B}}} & \lbrack 150\rbrack\end{matrix}$

Then the following inequality holds at h in an interval included betweenα and β.

f ₁(h)<0  [151]

Graph y=f₂(h) passes through the origin, h=−1/μ and h=−(δ²+φ²)/δ. Graphsy=f₁(h) and y=F₂(h) are shown in FIG. 3. From FIG. 3, it can be seenthat the following inequality holds at h in an interval included betweenα and β.

f ₁(h)<f ₂(h)  [152]

If G is zero, then m(h)=f₁(h) and the roots of m(h)=0 coincide with theones of f₁(h)=0. So as G increases in a positive direction, then αdecreases and β increases in the absolute value.

In the case of that the equation f₁(h)=f₂(h) has two imaginary roots,the position of the imaginary roots influences the stability offeedback. The magnitude of β is restricted the position of the imaginaryroots. For example, in the case such as

−S>>8δ,  [153]

let β be equal to −2δ, the real part of the imaginary roots is estimatedin the neighborhood of −3δ, where α in the neighborhood of the origin isignored. So the roots of m(h)=0 lines up in the sequence of α, β and thereal part of the imaginary roots from the origin.

In the case that the output voltage is not fed back to the amplitude, β,the root of f₁(h)=0, located close to −2δ brings the real part of theimaginary root in the neighborhood of the origin, and causes the ringingof the output voltage. The feedback to the amplitude attenuates theringing.

In such a case that the feedback of the output voltage to the frequencyof the carrier is not valid as is described by the differential equationsystem defined by expressions 78˜80, the frequency of the carrier isfixed, while the resonance frequency of the resonance circuit depends onthe load. Therefore the load is limited by the available range of theresonance frequency. The feedback of the output voltage to the frequencyof the carrier removes the restrictions on the load.

Advantageous Effect of Invention

Feeding back the output voltage to both the amplitude and the frequencyof the carrier driving a resonance circuit improves the characteristicsof the power supply using the resonance circuit, making thecharacteristics approximate to the ones of the power supply using theconventional transformer.

BRIEF DESCRIPTION OF DRAWINGS

FIG. 1 is an explanation chart where resonance frequency and the rangeof the driving frequency were shown.

FIG. 2 is an explanation chart where a block diagram for a model offeedback is shown.

FIG. 3 is an explanation chart where roots of equation m_(A)(h)=0 areshown.

FIG. 4 is an explanation chart where a block diagram of the power supplyis shown.

FIG. 5 is an explanation chart where the circuit for simulation isshown.

FIG. 6 is an explanation chart where an equivalent circuit of thepiezoelectric transformer is shown.

FIG. 7 is an explanation chart where the equivalent circuit of the idealtransformer is shown.

FIG. 8 is an explanation chart where frequency dependence of thepiezoelectric transformer is shown.

FIG. 9 is an explanation chart where the circuit for simulating thefrequency dependence of the piezoelectric transformer is shown.

FIG. 10 is an explanation chart where the circuit for simulating theerror amplifier is shown.

FIG. 11 is an explanation chart where the circuit for simulating thefeedback of the amplitude is shown.

FIG. 12 is an explanation chart where the circuit for simulating avoltage-controlled switch is shown.

FIG. 13 is an explanation chart where the circuit for simulating a drivepulse circuit is shown.

FIG. 14 is an explanation chart where results of simulation are shown.

BEST MODE FOR CARRYING OUT THE INVENTION

A block diagram of a stabilized dc. voltage power supply, using apiezoelectric transformer as a resonance circuit and composed of avoltage generation and a feedback circuits, is shown as implementationof the invention. A circuit simulating the voltage power supply denotedin the block diagram is given, showing the feedback to be stable.

[Mode for Invention 1]

The block diagram of the stabilized dc. voltage power supply is denotedin FIG. 4, where the a driver circuit, the piezoelectric transformer andthe rectification circuit compose the voltage generation circuit and thefeedback circuit consists of an error amplifier, a voltage controlledoscillator (VCO) and an amplitude compensation circuit.

Piezoelectric Transformer

The piezoelectric transformer utilizes piezoelectricity incorporated ina piezoelectric element. Stress applied to the element causes strainwhich gets voltage generated, and voltage applied to the element causesstress which gets strain produced. Utilizing the effect, thepiezoelectric transformer converts electrical vibration to mechanicalone at the primary, and restores the electrical vibration from themechanical one at the secondary, thus transmitting electric energy fromthe primary to the secondary. Transformation from the primary to thesecondary is attained through conversion between the electric energy andthe mechanical energy. The secondary of the transformer is capacitanceand the voltage is generated by the charge injected by the mechanicalvibration.

The piezoelectric transformer includes the resonance circuit. Therefore,the piezoelectric transformer shows sharp frequency characteristic andlarge load dependency, being different from the usual electromagnetictransformer. The piezoelectric transformer is used in the voltagegeneration circuit. Let amplitude ratio of the piezoelectric transformerbe defined by the ratio of the input to the output in voltage where thepiezoelectric transformer is connected to the load resistance, theamplitude ratio shows resonance characteristics as a function of thefrequency of the carrier.

Rectification Circuit

The output of the piezoelectric transformer is the carrier modulated inamplitude and rectified by a diode bridge. The output of the diodebridge is buffered by a capacitance. The capacitance reduces the voltageripples in the output voltage. While the output voltage is positive, therectification circuit can raise the output voltage by pumping up thecharge into the capacitance, but cannot lower the output voltage bypumping the charge out of the capacitance. A portion of the chargestored in the transformer is pumped up in a cycle of the carrierfrequency.

Considering the change of the output voltage in the case of the positiveoutput voltage, the rise of the output voltage makes the piezoelectrictransformer drive the load and charge up the capacitance. The outputvoltage is lowered through discharge of the capacitance through theload. Then the time constant for the rise become shorter as the loadbecomes lighter, while the output voltage falls at the time constantcaused by the capacitance and the load.

Error Amplifier

The error amplifier is composed of the subtraction circuit and the phasecompensation circuit. The subtraction circuit outputs voltage differencebetween the output voltage and the reference voltage.

The output of the subtraction circuit is supplied to the phasecompensation circuit which is composed of an amplifier and its feedbackbranch. The transfer function of the phase compensation circuit isprovided with a pole located at the origin and two zeros. The transferfunction converts the input to the output of the phase compensationcircuit. The output is clamped by diodes within in a limited range ofvoltage.

Amplitude Compensation Circuit

The output of the subtraction circuit in the error amplifier is suppliedto the amplitude compensation circuit. The amplitude compensationcircuit cancels the delay caused by the rectification circuit. Theoutput is supplied to a drive pulse circuit.

Driver Circuit

Viewed from the input, the piezoelectric transformer looks capacitive.The sinusoidal carrier is indispensable to drive the capacitanceefficiently, and the approximate sinusoidal carrier is generated byresonating with inductance.

The driver circuit driving the piezoelectric transformer is composed ofa full bridge including a pair of half bridges. The half bridgecomprises two FETs and a circuit to drive the FETs. Two FETs in the halfbridge are driven by a pair of complementary drive pulses. Two halfbridges composing the driver circuit are driven by four drive pulses,which are of the same frequency and grouped into two pairs of thecomplementary drive pulses.

The full bridge in the driver circuit is operated in a phase-shift mode.Four FETs composing the full bridges are driven by two pairs of thedrive pulses. There are phase shift between the pairs. The amplitude ofthe carrier is controlled by the phase shift between the pairs. Thefrequency of the drive pulses are generated by the VCO, and the phaseshift is controlled by the amplitude compensation circuit. A drive pulsecircuit generates the drive pulses.

Drive Pulse Circuit

The output of the phase compensation circuit and the one of theamplitude compensation circuit are both supplied to the drive pulsecircuit. The drive pulse circuit includes the VCO, to which the outputof the phase compensation circuit is supplied. The VCO produces a pulsesequence whose frequency is proportional to the input voltage. Ongenerating two pairs of the drive pulses, the amplitude compensationcircuit control the phase shift between the pairs.

Auxiliary Power Supply

The auxiliary power supply provides a dc. voltage to the driver circuitgenerating the carrier to drive the resonance circuit.

Circuit Simulating Power Supply

FIG. 5 shows the circuit simulating the stabilized dc. voltage powersupply, where the feedback is shown to be stable by simulation. Thecircuit simulating the voltage generation circuit is faithfulreproduction of an actual circuit except that the piezoelectrictransformer is replaced with an equivalent circuit. Fundamentally thefeedback circuit is linearly related between its input and output. So inthe circuit for simulation, the feedback circuit is replaced with asimple circuit reproducing the relation between the input and theoutput.

Equivalent Circuit for Piezoelectric Transformer

The equivalent circuit of the piezoelectric transformer used in thepower supply and its parameters are shown in FIG. 6. The equivalentcircuit includes an ideal transformer. Let n be the winding ratiobetween primary and secondary coils of the ideal transformer shown inFIG. 7. Let the primary voltage and current, and the secondary voltageand current be denoted by E₁, I₁, E₂ and I₂ respectively, then

$\begin{matrix}{\frac{E_{1}}{E_{2}} = {\frac{I_{2}}{I_{1}} = n}} & \lbrack 154\rbrack\end{matrix}$

Frequency Dependence of Piezoelectric Transformer

In FIG. 8, the frequency dependence of the piezoelectric transformer isshown where the load resistance is changed from 1Ω to 200 mΩ with adecrement of 200 mΩ. The circuit simulating the frequency dependence isrepresented by FIG. 9, where the load resistance is denoted by R76.

Circuit Simulating Error Amplifier

A circuit element called a behavior model, where the relation betweenthe input and the output is defined by a mathematical expression, isavailable for simulation. The circuit simulating the error amplifier isshown in FIG. 10. The error amplifier is composed of the subtractioncircuit, phase compensation circuit and a first order delay. Thesubtraction circuit is constructed of the behavior model provided withtwo input and an output terminals, where the output is equated withvoltage difference between the input. The output of the behavior modelis amplified by a gain block and supplied to the circuit simulating thephase compensation circuit.

The circuit simulating the phase compensation circuit is composed of anintegral part, a proportional part, a differential part, and two adders.The proportional part is composed of a gain block of unity gain. Theintegral part consists of a gain block and a behavior model. Thebehavior model provided with function SDT(X) outputs the integration ofthe input. The output is amplified by the gain block. The output of thegain block is the output of the integration part.

The differential part is composed of a gain block and a behavior model.The behavior model provided with function DDT(x) outputs thedifferentiation of the input relative to time. The output of thebehavior model is amplified by the gain block. The output of the gainblock is the output of the differential part.

Let E be the gain of the integration part and B be the one of thedifferential part, then the transfer function of the error amplifier isgiven by

$\begin{matrix}{\frac{E}{s} + 1 + {B\; {s.}}} & \lbrack 155\rbrack\end{matrix}$

The first order delay is defined by

$\begin{matrix}\frac{1}{1 + {A\; s}} & \lbrack 156\rbrack\end{matrix}$

where A is the time constant of the delay. A Laplace element generatingthe delay is available for simulation. The first order delay isimplemented by the Laplace element.

In FIG. 5, element S is the voltage source generating the referencevoltage. It is possible to control the loop gain by C. In the figure,E=15, B=0.0001 and C=1.

Circuit Simulating Amplitude Compensation Circuit

The circuit simulating the amplitude compensation circuit is shown inFIG. 11. The circuit for the compensation is composed of the circuitcompensating the first order delay and a low-pass filter. The circuitfor the compensation cancels the delay

$\begin{matrix}\frac{1}{1 + {\mu \; s}} & \lbrack 157\rbrack\end{matrix}$

caused by the rectification circuit. Then the transfer function of thecircuit for the compensation approximates the one given by

1+μs.  [158]

The circuit for the compensation is composed of a gain block of unitygain, a differentiator and an adder with two input terminals. Thelow-pass filter attenuates frequency components outside the bandwidthwhere the feedback is effective. The low-pass filter is of the firstorder delay and implemented by the Laplace element.

Circuit Simulating Driver Circuit

The circuit simulating the driver circuit is composed ofvoltage-controlled switches and the drive pulse circuit. A FET used inactual circuits is simulated by a four-terminal voltage-controlledswitch shown in FIG. 12. Terminals 1 and 2 of the switch are the controlinput, and the voltage difference being higher than 1 V between themmakes the switch conduct between terminals 3 and 4. The resistance ofthe switch is 0.1Ω while the switch is turned on, otherwise theresistance is 1 M Ω.

The driver circuit includes 4 voltage controlled switches A, B, C and D,where A and B composes one half bridge, and C and D composes the otherone. Drive pulses a, b, c and d are generated by a drive pulse circuit.The drive pulses a and b are complementary, driving A and B of the halfbridge individually. Similarly drive pulses c and d drive C and Dindividually.

Circuit Simulating Drive Pulse Circuit

Circuit simulating the drive pulse circuit is shown in FIG. 13. In orderto ease simulation, the circuit for simulation, being different from anactual circuit, is constructed on a basis of mathematical relation. Theoutput of a frequency modulation circuit included in the drive pulsecircuit is not of a pulse sequences but of sinusoidal wave. Thecomplementary drive pulses provided with dead time are generated byclamping the sinusoidal wave with an appropriate dc. voltage level.

The drive pulse circuit includes two frequency modulation circuits A andB. The sinusoidal wave supplied by B is shifted in phase relative to thesinusoidal wave supplied by A. The amount of the phase shift iscontrolled by the input of the frequency modulation cir cut,equivalently the output of the amplitude compensation circuit.

The output of the amplitude compensation circuit shifts the phase of thesinusoidal wave supplied by the circuit B relative to the one of thecircuit A between 0 and π.

Circuit Simulating Frequency Modulation Circuit

The circuit simulating the frequency modulation circuit is composed oftwo behavior models M1 and M2 as shown in FIG. 13. M outputs theintegration of its input. M2 outputs the sinusoidal wave with the phasespecified by its input. As a result, the M2 supplies the sinusoidal wavewith the frequency proportional to the input of M1, namely the output ofthe error amplifier.

Simulation Example of Stable Feedback

The circuit for simulation in FIG. 5 is utilized to show that thefeedback implemented in the stabilized dc. voltage power supply isstable. The rated voltage of the power supply is 1 V. The output voltagetogether with the input of the frequency modulation circuit and theoutput of the subtraction circuit are shown in FIG. 14 when the outputcurrent is added by 600 mA for 100 msec to the stationary output currentof 1 A. In the figure, a horizontal axis shows time, and vertical axes1, 2 and 3 correspond to the input of the frequency modulation circuit,the output of the subtraction circuit and the output voltagerespectively.

1. At a resonance circuit, letting ω_(r), Q and g_(r) be the angular velocity of the resonance frequency, the Q-value and the amplitude ratio at the resonance frequency respectively of resonance, then δ, ω₀ and c are defined by $\begin{matrix} {\delta = \frac{\omega_{r}}{2Q}} & \lbrack 1\rbrack \\ {\omega_{0} = {\omega_{r}\sqrt{1 - \frac{1}{4Q^{2}}}}} & \lbrack 2\rbrack \\ {c = \frac{g_{r}\omega_{r}}{Q}} & \lbrack 3\rbrack \end{matrix}$ and the resonance circuit is driven by a carrier of variable amplitude modulated in frequency where, letting w be a positive constant and x be a function of time, the amplitude is expressed by w+x  [4] and, letting ψ be phase in a function of time, the carrier is defined by (w+x)exp(iω₀t+iψ)  [5] then, approximating the transfer function of the resonance circuit, where the bandwidth for frequency modulation is narrow enough compared with the resonance frequency, by $\begin{matrix} {\frac{1}{2{\omega}_{0}}\frac{c\left( {{- \delta} + {\omega}_{0}} \right)}{s + \delta - {\omega}_{0}}} & \lbrack 6\rbrack \end{matrix}$ and defining a map from ψ in expression 5 to φ as $\begin{matrix} {\varphi = {\frac{}{t}\psi}} & \lbrack 7\rbrack \end{matrix}$ then the frequency of the carrier given in expression 5 is rewritten to ω₀+φ  [8] and, letting r_(r) and r_(i) be defined by $\begin{matrix} {r_{r} = {\frac{1}{2}c}} & \lbrack 9\rbrack \\ {r_{i} = {\frac{\delta}{2\omega_{0}}c}} & \lbrack 10\rbrack \end{matrix}$ the resonance circuit, the transfer function of which is given in expression 6, being driven by the carrier in expression 5, outputs the carrier the amplitude of which is √{square root over (p²+q²)}  [11] where p and q satisfy equations given by $\begin{matrix} {{\frac{}{t}p} = {{q\; \varphi} - {p\; \delta} + {r_{r}\left( {w + x} \right)}}} & \lbrack 12\rbrack \\ {{\frac{}{t}q} = {{{- p}\; \varphi} - {q\; \delta} + {r_{i}\left( {w + x} \right)}}} & \lbrack 13\rbrack \end{matrix}$ and rectification and smoothing cause a first order delay which, letting the dc. voltage generated by rectifying the carrier outputted by the resonance circuit be z, is represented by the following differential equation concerning z $\begin{matrix} {{{\mu \frac{}{t}z} + z} = {\nu \sqrt{p^{2} + q^{2}}}} & \lbrack 14\rbrack \end{matrix}$ where μ and ν are a time constant and a multiplier at a rectification circuit. Letting G and H be positive numbers and λ be a reference voltage respectively, the dc. voltage z in expression 14 from the rectification circuit is compared with the reference voltage λ and the voltage difference between z and λ is fed back to the amplitude of the carrier w+x in expression 5 as $\begin{matrix} {x = {{- {G\left( {z - \lambda} \right)}} - {H\frac{z}{t}}}} & \lbrack 15\rbrack \end{matrix}$ and substituting expression 15 for x in expressions 12 and 13 leads to $\begin{matrix} {{\frac{}{t}p} = {{\varphi \; q} - {\delta \; p} + {\left( {w - {G\left( {z - \lambda} \right)} - \frac{H\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu}} \right)r_{r}}}} & \lbrack 16\rbrack \\ {{\frac{}{t}q} = {{{- \varphi}\; p} - {\delta \; q} + {\left( {w - {G\left( {z - \lambda} \right)} - \frac{H\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu}} \right)r_{i}}}} & \lbrack 17\rbrack \end{matrix}$ and, letting k, d, E, A and B be positive numbers and λ be a reference voltage respectively, the dc. voltage z in expression 14 from the rectification circuit is compared with the reference voltage λ and the voltage difference between z and λ is fed back to the frequency of the carrier φ in expression 7, which is expressed, on the assumption that transfer function is provided with a pole located at the origin, as $\begin{matrix} {\varphi = {{kd}\frac{\left( {E + {As} + {Bs}^{2}} \right)}{s}\left( {z - \lambda} \right)}} & \lbrack 18\rbrack \end{matrix}$ then uniting expression 16, expressions 17 and expression 14 with expression 18 makes the system of differential equations as $\begin{matrix} {\mspace{79mu} {{\frac{}{t}p} = {{\varphi \; q} - {\delta \; p} + {\left( {w - {G\left( {z - \lambda} \right)} - \frac{H\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu}} \right)r_{r}}}}} & \lbrack 19\rbrack \\ {\mspace{79mu} {{\frac{}{t}q} = {{{- \varphi}\; p} - {\delta \; q} + {\left( {w - {G\left( {z - \lambda} \right)} - \frac{H\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu}} \right)r_{i}}}}} & \lbrack 20\rbrack \\ {\mspace{79mu} {{\frac{}{t}z} = \frac{{- z} + {v\sqrt{p^{2} + q^{2}}}}{\mu}}} & \lbrack 21\rbrack \\ {{\frac{}{t}\varphi} = {{- {{kdE}\left( {{- z} + \lambda} \right)}} + \frac{{Akd}\left( {{- z} + {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu} + {B\left( {{- \frac{{kd}\left( {{- z} + {v\sqrt{p^{2} + q^{2}}} + {\sqrt{p^{2} + q^{2}}v\; {\mu\delta}}} \right)}{\mu^{2}}} + \frac{{kdvw}\left( {{pr}_{r} + {r_{i}q}} \right)}{\mu \sqrt{p^{2} + q^{2}}}} \right)} + \frac{{{GkdBv}\left( {{pr}_{r} + {r_{i}q}} \right)}\left( {{- z} + \lambda} \right)}{\mu \sqrt{p^{2} + q^{2}}} + \frac{{{HkdBv}\left( {{pr}_{r} + {r_{i}q}} \right)}\left( {z - {v\sqrt{p^{2} + q^{2}}}} \right)}{\mu^{2}\sqrt{p^{2} + q^{2}}}}} & \lbrack 22\rbrack \end{matrix}$ Letting p_(e), q_(e), z_(e), φ_(e) be the equilibrium point of the system of differential equations, then z_(e)=λ and equivalently x=0. It can be seen that at the equilibrium point the amplitude and the frequency of the carrier are w and φ_(e) respectively provided that the amplitude of w is enough to generate the output voltage z_(e). Let r be defined by r=√{square root over (r_(r) ² +r _(i) ²)}  [23] then p_(e), q_(e), and z_(e) are given in functions of φ_(e) as $\begin{matrix} {p_{e} = \frac{w\left( {{\delta \; r_{r}} + {r_{i}\varphi_{e}}} \right)}{\varphi_{e}^{2} + \delta^{2}}} & \lbrack 24\rbrack \\ {q_{e} = {- \frac{w\left( {{{- \delta}\; r_{i}} + {\varphi_{e}r_{r}}} \right)}{\varphi_{e}^{2} + \delta^{2}}}} & \lbrack 25\rbrack \\ {z_{e} = \frac{vwr}{\sqrt{\varphi_{e}^{2} + \delta^{2}}}} & \lbrack 26\rbrack \\ {\lambda = \frac{vwr}{\sqrt{\varphi_{e}^{2} + \delta^{2}}}} & \lbrack 27\rbrack \end{matrix}$ Letting m(h) be the characteristic polynomial of the differential equation system approximating the above system of differential equations linearly in the neighborhood of the equilibrium point, and letting m(h) be expressed as m(h)=a ₀ h ⁴ +a ₁ h ³ +a ₂ h ² +a ₃ h+a ₄  [28] coefficients a₀, a₁, a₂, a₃, and a₄ are given as $\begin{matrix} {a_{0} = 1} & \lbrack 29\rbrack \\ {a_{1} = \frac{{\delta \; {vHr}} + {2\; {\delta\mu}\sqrt{\varphi_{e}^{2} + \delta^{2}}} + \sqrt{\varphi_{e}^{2} + \delta^{2}}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 30\rbrack \\ {a_{2} = \frac{\begin{matrix} {{2\delta \sqrt{\varphi_{e}^{2} + \delta^{2}}} + {\delta \; {vrG}} + {\varphi_{e}{kdvwrB}} +} \\ {{\mu \left( {\varphi_{e}^{2} + \delta^{2}} \right)^{3/2}} + {{rHv}\left( {\varphi_{e}^{2} + \delta^{2}} \right)}} \end{matrix}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 31\rbrack \\ {a_{3} = \frac{{\varphi_{e}{wrkdvA}} + \left( {\varphi_{e}^{2} + \delta^{2}} \right)^{3/2} + {{vrG}\left( {\varphi_{e}^{2} + \delta^{2}} \right)}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 32\rbrack \\ {a_{4} = \frac{\varphi_{e}{kdvwrE}}{\sqrt{\varphi_{e}^{2} + \delta^{2}}\mu}} & \lbrack 33\rbrack \end{matrix}$ where r be defined by r=√{square root over (r_(r) ² +r _(i) ²)}  [34] Hereafter φ_(e) is replaced with φ for simplicity, the characteristic polynomial m(h) is represented by $\begin{matrix} {{m(h)} = {\frac{{h\left( {\delta^{2} + {2h\; \delta} + h^{2} + \varphi^{2}} \right)}\left( {{h\; \mu} + 1} \right)}{\mu} + \frac{\varphi \; {{wk}\left( {E + {hA} + {h^{2}B}} \right)}{vrd}}{\sqrt{\varphi^{2} + \delta^{2}}\mu} + \frac{{{hvr}\left( {\delta^{2} + {h\; \delta} + \varphi^{2}} \right)}\left( {G + {hH}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}} & \lbrack 35\rbrack \end{matrix}$ Letting F_(w) and B_(w) be the transfer functions of the voltage generation circuit and the feedback circuit respectively, the characteristic polynomial m(h) coincides with the denominator of closed loop transfer function F_(w)/(1+F_(w)B_(k)) except for a constant multiplier, and stable feedback in the sense that all the roots of the denominator of the closed loop transfer function has a negative real part is realized by selecting E, B, and N to be such that all the roots of the characteristic polynomial have a negative real part
 2. Letting υ be the output voltage of the subtraction circuit, the voltage υ is the voltage difference between the output voltage z and the reference voltage λ, and υ is fed back to the frequency and the amplitude of the carrier, building up the output voltage. Let the part of the output voltage which υ contributes through the frequency be denoted by fυ  [36] and the part of the output voltage due to the amplitude is defined by gυ  [37] then the output voltage z is expressed by z=fυ+gυ  [38] and at the subtraction circuit υ is written as υ=λ−z  [39] hence z is given by $\begin{matrix} {z = \frac{\left( {f + g} \right)\lambda}{1 + f + g}} & \lbrack 40\rbrack \end{matrix}$ where the equation 0=1+f+g  [41] is equivalent to the equation 0=m(h)  [42] where m(h) is the characteristic polynomial, then f and g can be found except for a common constant as $\begin{matrix} {f = \frac{k\; \varphi \; {w\left( {E + {Ah} + {Bh}^{2}} \right)}}{\sqrt{\varphi^{2} + \delta^{2}}{h\left( {\varphi^{2} + h^{2} + \delta^{2} + {2\; h\; \delta}} \right)}\left( {1 + {h\; \mu}} \right)}} & \lbrack 43\rbrack \\ {g = \frac{{{vr}\left( {\varphi^{2} + {h\; \delta} + \delta^{2}} \right)}\left( {G + {hH}} \right)}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + h^{2} + \delta^{2} + {2\; h\; \delta}} \right)\left( {1 + {h\; \mu}} \right)}} & \lbrack 44\rbrack \end{matrix}$ Let μ₁ be selected as μ₁˜μ  [45] and G be related with H as H=μ₁G  [46] where μ₁ and G are positive constants, then g can be rewritten as $\begin{matrix} {g = {\frac{{{vr}\left( {\varphi^{2} + \delta^{2} + {h\; \delta}} \right)}{G\left( {{\mu_{1}h} + 1} \right)}}{\sqrt{\varphi^{2} + \delta^{2}}\left( {\varphi^{2} + \delta^{2} + h^{2} + {2\; h\; \delta}} \right)\left( {{h\; \mu} + 1} \right)}.}} & \lbrack 47\rbrack \end{matrix}$ As for a term E+Ah+Bh² in the numerator of f in expression 43, it is assumed that the following equation of the second degree in h E+Ah+Bh ²=0  [48] has such a root that is close to cancel the term 1+μh in the denominator. Let f₁(h) and f₂(h) be defined by $\begin{matrix} {{f_{1}(h)} = {\frac{{h\left( {\delta^{2} + {2\; h\; \delta} + h^{2} + \varphi^{2}} \right)}\left( {{h\; \mu} + 1} \right)}{\mu} + \frac{{kw}\; {\varphi \left( {{hA} + {h^{2}B} + E} \right)}{vrd}}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}} & \lbrack 49\rbrack \\ {\mspace{79mu} {{{f_{2}(h)} = {- \frac{{{hvr}\left( {\delta^{2} + {h\; \delta} + \varphi^{2}} \right)}{G\left( {1 + {h\; \mu_{1}}} \right)}}{\sqrt{\varphi^{2} + \delta^{2}}\mu}}}\mspace{79mu} {then}}} & \lbrack 50\rbrack \\ {\mspace{79mu} {{m(h)} = {{f_{1}(h)} - {f_{2}(h)}}}} & \lbrack 51\rbrack \end{matrix}$ and letting m(h) be expressed by m(h)=a ₀ h ⁴ +a ₁ h ³ +a ₂ h ² +a ₃ h+a ₄  [52] then the coefficients a₀ and a₁ are given by $\begin{matrix} {a_{0} = 1} & \lbrack 53\rbrack \\ {a_{1} = {{\frac{\delta \; {vHr}}{\sqrt{\varphi^{2} + \delta^{2}}\mu} + \frac{{2{\delta\mu}} + 1}{\mu}} = {\frac{\delta \; {vG}\; \mu_{1}r}{\sqrt{\varphi^{2} + \delta^{2}}\mu} + \frac{{2{\delta\mu}} + 1}{\mu}}}} & \lbrack 54\rbrack \end{matrix}$ and μ₁ is approximate to μ and from expressions 26 and 27, the expression $\begin{matrix} \frac{vwr}{\sqrt{\varphi_{e}^{2} + \delta^{2}}} & \lbrack 55\rbrack \end{matrix}$ is equal to the reference voltage λ and z, and G is roughly estimated that $\begin{matrix} {\left\lbrack \left\lbrack {\frac{G}{w}1} \right\rbrack \right\rbrack {\frac{G}{w} \sim 1.}} & \lbrack 56\rbrack \end{matrix}$ Anyway, letting the sum of the root of m(h)=0 be S, S is estimated as −S≧2δ  [57] The characteristic polynomial of the feedback where, the amplitude of the carrier being fixed and equal to w, the output voltage is fed back to the frequency of the carrier is equal to f₁(h), and letting α be the real characteristic root of the equation f₁(h)=0, which is locating between the origin and −1/μ, and letting β be the real root next to α, the feedback is stabilized in the case that G=0 by the appropriate arrangement of α and β, where α and β are roughly estimated by the coefficients of the term E+As+Bs² in f₁(h) as $\begin{matrix} {\alpha \sim {- \frac{E}{A}}} & \lbrack 58\rbrack \\ {\beta \sim {- \frac{A}{2\; B}}} & \lbrack 59\rbrack \end{matrix}$ then there exists an interval included between α and β such that h in the interval satisfies the inequality f ₁(h)<0,  [60] and the graph of y=f₂(h) passes through the origin, −1/μ and h=−(δ²+φ²)δ, and then there is an interval included between α and β where h in the interval satisfies the inequality f ₁(h)<f ₂(h).  [61] The roots of m(h)=0 coincides with the roots of f₁(h)=0 in the case that G=0 and as G grows in the positive direction, S increases in the absolute value, which can move β in a negative direction keeping the real parts of the roots negative. Furthermore in the case that the equation f₁(h)=f₂(h) has imaginary roots, letting γ be the real part of the imaginary roots, then γ is related to α, β and S by α+β+2γ=S  [62] by which the real part of the imaginary root can be controlled through E, A, B, G, and H so as to realize the stable feedback.
 3. A voltage source comprises a voltage generation circuit and a feedback circuit. The voltage generation circuit is composed of a driver circuit, the resonance circuit, and the rectification circuit. The driver circuit generates e high-frequency alternating current carrier of variable amplitude. The carrier drives the resonance circuit. The output of the resonance circuit is rectified by the rectification circuit to be a dc. voltage which is the output voltage of the voltage source. The feedback circuit consists of a subtraction circuit, a phase compensation circuit and an amplitude compensation circuit which are supplied with the output of the subtraction circuit, where the subtraction circuit outputs the voltage difference between the output voltage and the reference voltage supplied externally. The frequency compensation circuit has a mean to control a voltage controlled oscillator in the driver circuit. The amplitude compensation circuit has a mean to control an amplitude modulator in the driver circuit. The carrier generated by the driver circuit is modulated both in frequency and in amplitude by the output of the subtraction circuit, and then the resonance circuit outputs the carrier modulated in amplitude by the output of the subtraction circuit. The carrier is demodulated by the rectification circuit to be the output voltage approximating the output of the subtraction circuit. Letting υ be the output voltage of the subtraction circuit and z be the output voltage, let the part of the output voltage which υ contributes through the frequency be denoted by f υ and the part of the output voltage contributed through the amplitude be defined by g υ, then z=fυ+gυ  [63] Letting the transfer function of the frequency compensation circuit be f_(B) and the transfer function of the amplitude compensation circuit be g_(B), f and g can be expressed as f=f_(F)f_(B)  [64] g=g_(F)g_(B)  [65] where the forward transfer functions f_(F) and g_(F) share the common term originated by the delay caused by the rectification circuit. Canceling the delay due to the common term approximately by f_(B) and g_(B), and the frequency feedback being larger than amplitude feedback in delay, eliminating interference between the frequency and the amplitude feedback by locating the zero of the frequency compensation circuit in the neighborhood of the origin establish stable feedback where the amplitude feedback with wide-bandwidth and high-gain is implemented together with the frequency feedback. 